Section 2.1: Linear Functions, from College Algebra: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a Creative

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1 Section.: Linear Functions, from College Algebra: Corrected Edition b Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a Creative Commons Attribution-NonCommercial-ShareAlike.0 license. 0, Carl Stitz.

2 Chapter Linear and Quadratic Functions. Linear Functions We now begin the stud of families of functions. Our first famil, linear functions, are old friends as we shall soon see. Recall from Geometr that two distinct points in the plane determine a unique line containing those points, as indicated below. P ( 0, 0 ) Q (, ) To give a sense of the steepness of the line, we recall that we can compute the slope of the line using the formula below. Equation.. The slope m of the line containing the points P ( 0, 0 ) and Q (, ) is: provided 0. m = 0 0, A couple of notes about Equation. are in order. First, don t ask wh we use the letter m to represent slope. There are man eplanations out there, but apparentl no one reall knows for sure. Secondl, the stipulation 0 ensures that we aren t tring to divide b zero. The reader is invited to pause to think about what is happening geometricall; the anious reader can skip along to the net eample. Eample... Find the slope of the line containing the following pairs of points, if it eists. Plot each pair of points and the line containing them. See or for discussions on this topic.

3 5 Linear and Quadratic Functions. P (0, 0), Q(, ). P (, ), Q(, ). P (, ), Q(, ). P (, ), Q(, ) 5. P (, ), Q(, ) 6. P (, ), Q(., ) Solution. In each of these eamples, we appl the slope formula, Equation... m = 0 0 = = Q P. m = ( ) = = P Q P. m = ( ) = 6 = Q. m = ( ) = 0 7 = 0 P Q

4 . Linear Functions 5 P 5. m = =, which is undefined 0 Q P 6. m =. = 0. = 0 Q A few comments about Eample.. are in order. First, for reasons which will be made clear soon, if the slope is positive then the resulting line is said to be increasing. If it is negative, we sa the line is decreasing. A slope of 0 results in a horizontal line which we sa is constant, and an undefined slope results in a vertical line. Second, the larger the slope is in absolute value, the steeper the line. You ma recall from Intermediate Algebra that slope can be described as the ratio rise run. For eample, in the second part of Eample.., we found the slope to be. We can interpret this as a rise of unit upward for ever units to the right we travel along the line, as shown below. over up Some authors use the unfortunate moniker no slope when a slope is undefined. It s eas to confuse the notions of no slope with slope of 0. For this reason, we will describe slopes of vertical lines as undefined.

5 5 Linear and Quadratic Functions Using more formal notation, given points ( 0, 0 ) and (, ), we use the Greek letter delta to write = 0 and = 0. In most scientific circles, the smbol means change in. Hence, we ma write m =, which describes the slope as the rate of change of with respect to. Rates of change abound in the real world, as the net eample illustrates. Eample... Suppose that two separate temperature readings were taken at the ranger station on the top of Mt. Sasquatch: at 6 AM the temperature was F and at 0 AM it was F.. Find the slope of the line containing the points (6, ) and (0, ).. Interpret our answer to the first part in terms of temperature and time.. Predict the temperature at noon. Solution.. For the slope, we have m = 0 6 = 8 =.. Since the values in the numerator correspond to the temperatures in F, and the values in the denominator correspond to time in hours, we can interpret the slope as = = F hour, or F per hour. Since the slope is positive, we know this corresponds to an increasing line. Hence, the temperature is increasing at a rate of F per hour.. Noon is two hours after 0 AM. Assuming a temperature increase of F per hour, in two hours the temperature should rise F. Since the temperature at 0 AM is F, we would epect the temperature at noon to be + = 6 F. Now it ma well happen that in the previous scenario, at noon the temperature is onl F. This doesn t mean our calculations are incorrect, rather, it means that the temperature change throughout the da isn t a constant F per hour. As discussed in Section.., mathematical models are just that: models. The predictions we get out of the models ma be mathematicall accurate, but ma not resemble what happens in the real world. In Section., we discussed the equations of vertical and horizontal lines. Using the concept of slope, we can develop equations for the other varieties of lines. Suppose a line has a slope of m and contains the point ( 0, 0 ). Suppose (, ) is another point on the line, as indicated below. (, ) ( 0, 0 )

6 . Linear Functions 55 Equation. ields m = 0 0 m ( 0 ) = 0 0 = m ( 0 ) We have just derived the point-slope form of a line. Equation.. The point-slope form of the line with slope m containing the point ( 0, 0 ) is the equation 0 = m ( 0 ). Eample... Write the equation of the line containing the points (, ) and (, ). Solution. In order to use Equation. we need to find the slope of the line in question so we use Equation. to get m = = ( ) =. We are spoiled for choice for a point ( 0, 0 ). We ll use (, ) and leave it to the reader to check that using (, ) results in the same equation. Substituting into the point-slope form of the line, we get 0 = m ( 0 ) = ( ( )) = ( + ) = = + 7. We can check our answer b showing that both (, ) and (, ) are on the graph of = + 7 algebraicall, as we did in Section... In simplifing the equation of the line in the previous eample, we produced another form of a line, the slope-intercept form. This is the familiar = m + b form ou have probabl seen in Intermediate Algebra. The intercept in slope-intercept comes from the fact that if we set = 0, we get = b. In other words, the -intercept of the line = m + b is (0, b). Equation.. The slope-intercept form of the line with slope m and -intercept (0, b) is the equation = m + b. Note that if we have slope m = 0, we get the equation = b which matches our formula for a horizontal line given in Section.. The formula given in Equation. can be used to describe all lines ecept vertical lines. All lines ecept vertical lines are functions (Wh is this?) so we have finall reached a good point to introduce linear functions. We can also understand this equation in terms of appling transformations to the function I() =. See the Eercises.

7 56 Linear and Quadratic Functions Definition.. A linear function is a function of the form f() = m + b, where m and b are real numbers with m 0. The domain of a linear function is (, ). For the case m = 0, we get f() = b. These are given their own classification. Definition.. A constant function is a function of the form f() = b, where b is real number. The domain of a constant function is (, ). Recall that to graph a function, f, we graph the equation = f(). Hence, the graph of a linear function is a line with slope m and -intercept (0, b); the graph of a constant function is a horizontal line (a line with slope m = 0) and a -intercept of (0, b). Now think back to Section.6., specificall Definition.0 concerning increasing, decreasing and constant functions. A line with positive slope was called an increasing line because a linear function with m > 0 is an increasing function. Similarl, a line with a negative slope was called a decreasing line because a linear function with m < 0 is a decreasing function. And horizontal lines were called constant because, well, we hope ou ve alread made the connection. Eample... Graph the following functions. Identif the slope and -intercept.. f() =. f() =. f() =. f() = Solution.. To graph f() =, we graph =. This is a horizontal line (m = 0) through (0, ).. The graph of f() = is the graph of the line =. Comparison of this equation with Equation. ields m = and b =. Hence, our slope is and our -intercept is (0, ). To get another point on the line, we can plot (, f()) = (, ).

8 . Linear Functions 57 f() = f() =. At first glance, the function f() = does not fit the form in Definition. but after some rearranging we get f() = = = +. We identif m = and b =. Hence, our graph is a line with a slope of and a -intercept of ( 0, ). Plotting an additional point, we can choose (, f()) to get (, ).. If we simplif the epression for f, we get f() = = ( )( + ) ( ) = +. If we were to state f() = +, we would be committing a sin of omission. Remember, to find the domain of a function, we do so before we simplif! In this case, f has big problems when =, and as such, the domain of f is (, ) (, ). To indicate this, we write f() = +,. So, ecept at =, we graph the line = +. The slope m = and the -intercept is (0, ). A second point on the graph is (, f()) = (, ). Since our function f is not defined at =, we put an open circle at the point that would be on the line = + when =, namel (, ). f() = f() =

9 58 Linear and Quadratic Functions The last two functions in the previous eample showcase some of the difficult in defining a linear function using the phrase of the form as in Definition., since some algebraic manipulations ma be needed to rewrite a given function to match the form. Keep in mind that the domains of linear and constant functions are all real numbers (, ), so while f() = simplified to a formula f() = +, f is not considered a linear function since its domain ecludes =. However, we would consider f() = + + to be a constant function since its domain is all real numbers (Can ou tell us wh?) and f() = + + = ( + ) ( + ) = The following eample uses linear functions to model some basic economic relationships. Eample..5. The cost C, in dollars, to produce PortaBo game sstems for a local retailer is given b C() = for 0.. Find and interpret C(0).. How man PortaBos can be produced for $5,000?. Eplain the significance of the restriction on the domain, 0.. Find and interpret C(0). 5. Find and interpret the slope of the graph of = C(). Solution.. To find C(0), we replace ever occurrence of with 0 in the formula for C() to get C(0) = 80(0) + 50 = 950. Since represents the number of PortaBos produced, and C() represents the cost, in dollars, C(0) = 950 means it costs $950 to produce 0 PortaBos for the local retailer.. To find how man PortaBos can be produced for $5,000, we solve C() = 5000, or = Solving, we get = = Since we can onl produce a whole number amount of PortaBos, we can produce 85 PortaBos for $5,000.. The restriction 0 is the applied domain, as discussed in Section... In this contet, represents the number of PortaBos produced. It makes no sense to produce a negative quantit of game sstems. 5 The similarit of this name to PortaJohn is deliberate. 5 Actuall, it makes no sense to produce a fractional part of a game sstem, either, as we saw in the previous part of this eample. This absurdit, however, seems quite forgivable in some tetbooks but not to us.

10 . Linear Functions 59. We find C(0) = 80(0) + 50 = 50. This means it costs $50 to produce 0 PortaBos. As mentioned on page 8, this is the fied, or start-up cost of this venture. 5. If we were to graph = C(), we would be graphing the portion of the line = for 0. We recognize the slope, m = 80. Like an slope, we can interpret this as a rate of change. Here, C() is the cost in dollars, while measures the number of PortaBos so m = = C = 80 = 80 = $80 PortaBo. In other words, the cost is increasing at a rate of $80 per PortaBo produced. This is often called the variable cost for this venture. The net eample asks us to find a linear function to model a related economic problem. Eample..6. The local retailer in Eample..5 has determined that the number of PortaBo game sstems sold in a week is related to the price p in dollars of each sstem. When the price was $0, 0 game sstems were sold in a week. When the sstems went on sale the following week, 0 sstems were sold at $90 a piece.. Find a linear function which fits this data. Use the weekl sales as the independent variable and the price p as the dependent variable.. Find a suitable applied domain.. Interpret the slope.. If the retailer wants to sell 50 PortaBos net week, what should the price be? 5. What would the weekl sales be if the price were set at $50 per sstem? Solution.. We recall from Section. the meaning of independent and dependent variable. Since is to be the independent variable, and p the dependent variable, we treat as the input variable and p as the output variable. Hence, we are looking for a function of the form p() = m + b. To determine m and b, we use the fact that 0 PortaBos were sold during the week when the price was 0 dollars and 0 units were sold when the price was 90 dollars. Using function notation, these two facts can be translated as p(0) = 0 and p(0) = 90. Since m represents the rate of change of p with respect to, we have m = p 90 0 = 0 0 = 0 0 =.5. We now have determined p() =.5 + b. To determine b, we can use our given data again. Using p(0) = 0, we substitute = 0 into p() =.5 + b and set the result equal to 0:.5(0) + b = 0. Solving, we get b = 50. Hence, we get p() = We can check our formula b computing p(0) and p(0) to see if we get 0 and 90, respectivel. You ma recall from page 8 that the function p() is called the price-demand (or simpl demand) function for this venture.

11 60 Linear and Quadratic Functions. To determine the applied domain, we look at the phsical constraints of the problem. Certainl, we can t sell a negative number of PortaBos, so 0. However, we also note that the slope of this linear function is negative, and as such, the price is decreasing as more units are sold. Thus another constraint on the price is p() 0. Solving results in.5 50 or 500 = Since represents the number of PortaBos sold in a week, we round down to 66. As a result, a reasonable applied domain for p is [0, 66].. The slope m =.5, once again, represents the rate of change of the price of a sstem with respect to weekl sales of PortaBos. Since the slope is negative, we have that the price is decreasing at a rate of $.50 per PortaBo sold. (Said differentl, ou can sell one more PortaBo for ever $.50 drop in price.). To determine the price which will move 50 PortaBos, we find p(50) =.5(50)+50 = 5. That is, the price would have to be $5. 5. If the price of a PortaBo were set at $50, we have p() = 50, or,.5+50 = 50. Solving, we get.5 = 00 or = This means ou would be able to sell 66 PortaBos a week if the price were $50 per sstem. Not all real-world phenomena can be modeled using linear functions. Nevertheless, it is possible to use the concept of slope to help analze non-linear functions using the following. Definition.. Let f be a function defined on the interval [a, b]. The average rate of change of f over [a, b] is defined as: f f(b) f(a) = b a Geometricall, if we have the graph of = f(), the average rate of change over [a, b] is the slope of the line which connects (a, f(a)) and (b, f(b)). This is called the secant line through these points. For that reason, some tetbooks use the notation m sec for the average rate of change of a function. Note that for a linear function m = m sec, or in other words, its rate of change over an interval is the same as its average rate of change. = f() (a, f(a)) (b, f(b)) The graph of = f() and its secant line through (a, f(a)) and (b, f(b)) The interested reader ma question the adjective average in the phrase average rate of change. In the figure above, we can see that the function changes wildl on [a, b], et the slope of the secant line onl captures a snapshot of the action at a and b. This situation is entirel analogous to the

12 . Linear Functions 6 average speed on a trip. Suppose it takes ou hours to travel 00 miles. Your average speed is 00 miles hours = 50 miles per hour. However, it is entirel possible that at the start of our journe, ou traveled 5 miles per hour, then sped up to 65 miles per hour, and so forth. The average rate of change is akin to our average speed on the trip. Your speedometer measures our speed at an one instant along the trip, our instantaneous rate of change, and this is one of the central themes of Calculus. 6 When interpreting rates of change, we interpret them the same wa we did slopes. In the contet of functions, it ma be helpful to think of the average rate of change as: change in outputs change in inputs Eample..7. Recall from page 8, the revenue from selling units at a price p per unit is given b the formula R = p. Suppose we are in the scenario of Eamples..5 and..6.. Find and simplif an epression for the weekl revenue R() as a function of weekl sales.. Find and interpret the average rate of change of R() over the interval [0, 50].. Find and interpret the average rate of change of R() as changes from 50 to 00 and compare that to our result in part.. Find and interpret the average rate of change of weekl revenue as weekl sales increase from 00 PortaBos to 50 PortaBos. Solution.. Since R = p, we substitute p() =.5+50 from Eample..6 to get R() = (.5+ 50) = Since we determined the price-demand function p() is restricted to 0 66, R() is restricted to these values of as well.. Using Definition., we get that the average rate of change is R = R(50) R(0) 50 0 = = 75. Interpreting this slope as we have in similar situations, we conclude that for ever additional PortaBo sold during a given week, the weekl revenue increases $75.. The wording of this part is slightl different than that in Definition., but its meaning is to find the average rate of change of R over the interval [50, 00]. To find this rate of change, we compute 6 Here we go again... R = R(00) R(50) = = 5.

13 6 Linear and Quadratic Functions In other words, for each additional PortaBo sold, the revenue increases b $5. Note that while the revenue is still increasing b selling more game sstems, we aren t getting as much of an increase as we did in part of this eample. (Can ou think of wh this would happen?). Translating the English to the mathematics, we are being asked to find the average rate of change of R over the interval [00, 50]. We find R = R(50) R(00) = = 5. This means that we are losing $5 dollars of weekl revenue for each additional PortaBo sold. (Can ou think wh this is possible?) We close this section with a new look at difference quotients which were first introduced in Section.. If we wish to compute the average rate of change of a function f over the interval [, + h], then we would have f f( + h) f() f( + h) f() = = ( + h) h As we have indicated, the rate of change of a function (average or otherwise) is of great importance in Calculus. 7 Also, we have the geometric interpretation of difference quotients which was promised to ou back on page 8 a difference quotient ields the slope of a secant line. 7 So we are not torturing ou with these for nothing.

14 . Linear Functions 6.. Eercises In Eercises - 0, find both the point-slope form and the slope-intercept form of the line with the given slope which passes through the given point.. m =, P (, ). m =, P ( 5, 8). m =, P ( 7, ). m =, P (, ) 5. m = 5, P (0, ) 6. m = 7, P (, ) 7. m = 0, P (, 7) 8. m =, P (0, ) 9. m = 5, P (, ) 0. m = 678, P (, ) In Eercises - 0, find the slope-intercept form of the line which passes through the given points.. P (0, 0), Q(, 5). P (, ), Q(, ). P (5, 0), Q(0, 8). P (, 5), Q(7, ) 5. P (, 5), Q(7, 5) 6. P (, 8), Q(5, 8) 7. P (, ( ), Q 5, 7 ) 8. P (, 7 ) (, Q, ) 9. P (, ), Q (, ) 0. P (, ), Q (, ) In Eercises - 6, graph the function. Find the slope, -intercept and -intercept, if an eist.. f() =. f() =. f() =. f() = 0 5. f() = + 6. f() = 7. Find all of the points on the line = + which are units from the point (, ). 8. Jeff can walk comfortabl at miles per hour. Find a linear function d that represents the total distance Jeff can walk in t hours, assuming he doesn t take an breaks. 9. Carl can stuff 6 envelopes per minute. Find a linear function E that represents the total number of envelopes Carl can stuff after t hours, assuming he doesn t take an breaks. 0. A landscaping compan charges $5 per cubic ard of mulch plus a deliver charge of $0. Find a linear function which computes the total cost C (in dollars) to deliver cubic ards of mulch.

15 6 Linear and Quadratic Functions. A plumber charges $50 for a service call plus $80 per hour. If she spends no longer than 8 hours a da at an one site, find a linear function that represents her total dail charges C (in dollars) as a function of time t (in hours) spent at an one given location.. A salesperson is paid $00 per week plus 5% commission on her weekl sales of dollars. Find a linear function that represents her total weekl pa, W (in dollars) in terms of. What must her weekl sales be in order for her to earn $75.00 for the week?. An on-demand publisher charges $.50 to print a 600 page book and $5.50 to print a 00 page book. Find a linear function which models the cost of a book C as a function of the number of pages p. Interpret the slope of the linear function and find and interpret C(0).. The Topolog Tai Compan charges $.50 for the first fifth of a mile and $0.5 for each additional fifth of a mile. Find a linear function which models the tai fare F as a function of the number of miles driven, m. Interpret the slope of the linear function and find and interpret F (0). 5. Water freezes at 0 Celsius and Fahrenheit and it boils at 00 C and F. (a) Find a linear function F that epresses temperature in the Fahrenheit scale in terms of degrees Celsius. Use this function to convert 0 C into Fahrenheit. (b) Find a linear function C that epresses temperature in the Celsius scale in terms of degrees Fahrenheit. Use this function to convert 0 F into Celsius. (c) Is there a temperature n such that F (n) = C(n)? 6. Legend has it that a bull Sasquatch in rut will howl approimatel 9 times per hour when it is 0 F outside and onl 5 times per hour if it s 70 F. Assuming that the number of howls per hour, N, can be represented b a linear function of temperature Fahrenheit, find the number of howls per hour he ll make when it s onl 0 F outside. What is the applied domain of this function? Wh? 7. Economic forces beond anone s control have changed the cost function for PortaBos to C() = Rework Eample..5 with this new cost function. 8. In response to the economic forces in Eercise 7 above, the local retailer sets the selling price of a PortaBo at $50. Remarkabl, 0 units were sold each week. When the sstems went on sale for $0, 0 units per week were sold. Rework Eamples..6 and..7 with this new data. What difficulties do ou encounter? 9. A local pizza store offers medium two-topping pizzas delivered for $6.00 per pizza plus a $.50 deliver charge per order. On weekends, the store runs a game da special: if si or more medium two-topping pizzas are ordered, the are $5.50 each with no deliver charge. Write a piecewise-defined linear function which calculates the cost C (in dollars) of p medium two-topping pizzas delivered during a weekend.

16 . Linear Functions A restaurant offers a buffet which costs $5 per person. For parties of 0 or more people, a group discount applies, and the cost is $.50 per person. Write a piecewise-defined linear function which calculates the total bill T of a part of n people who all choose the buffet.. A mobile plan charges a base monthl rate of $0 for the first 500 minutes of air time plus a charge of 5 for each additional minute. Write a piecewise-defined linear function which calculates the monthl cost C (in dollars) for using m minutes of air time. HINT: You ma want to revisit Eercise 7 in Section.. The local pet shop charges per cricket up to 00 crickets, and 0 per cricket thereafter. Write a piecewise-defined linear function which calculates the price P, in dollars, of purchasing c crickets.. The cross-section of a swimming pool is below. Write a piecewise-defined linear function which describes the depth of the pool, D (in feet) as a function of: (a) the distance (in feet) from the edge of the shallow end of the pool, d. (b) the distance (in feet) from the edge of the deep end of the pool, s. (c) Graph each of the functions in (a) and (b). Discuss with our classmates how to transform one into the other and how the relate to the diagram of the pool. d ft. 7 ft. s ft. ft. 8 ft. 0 ft. 5 ft. In Eercises - 9, compute the average rate of change of the function over the specified interval.. f() =, [, ] 5. f() =, [, 5] 6. f() =, [0, 6] 7. f() =, [, ] 8. f() = +, [5, 7] 9. f() = + 7, [, ]

17 66 Linear and Quadratic Functions In Eercises 50-5, compute the average rate of change of the given function over the interval [, + h]. Here we assume [, + h] is in the domain of the function. 50. f() = 5. f() = 5. f() = + 5. f() = The height of an object dropped from the roof of an eight stor building is modeled b: h(t) = 6t + 6, 0 t. Here, h is the height of the object off the ground in feet, t seconds after the object is dropped. Find and interpret the average rate of change of h over the interval [0, ]. 55. Using data from Bureau of Transportation Statistics, the average fuel econom F in miles per gallon for passenger cars in the US can be modeled b F (t) = t + 0.5t + 6, 0 t 8, where t is the number of ears since 980. Find and interpret the average rate of change of F over the interval [0, 8]. 56. The temperature T in degrees Fahrenheit t hours after 6 AM is given b: T (t) = t + 8t +, 0 t (a) Find and interpret T (), T (8) and T (). (b) Find and interpret the average rate of change of T over the interval [, 8]. (c) Find and interpret the average rate of change of T from t = 8 to t =. (d) Find and interpret the average rate of temperature change between 0 AM and 6 PM. 57. Suppose C() = represents the costs, in hundreds, to produce thousand pens. Find and interpret the average rate of change as production is increased from making 000 to 5000 pens. 58. With the help of our classmates find several other real-world eamples of rates of change that are used to describe non-linear phenomena. (Parallel Lines) Recall from Intermediate Algebra that parallel lines have the same slope. (Please note that two vertical lines are also parallel to one another even though the have an undefined slope.) In Eercises 59-6, ou are given a line and a point which is not on that line. Find the line parallel to the given line which passes through the given point. 59. = +, P (0, 0) 60. = 6 + 5, P (, )

18 . Linear Functions = 7, P (6, 0) 6. =, P (, ) 6. = 6, P (, ) 6. =, P ( 5, 0) (Perpendicular Lines) Recall from Intermediate Algebra that two non-vertical lines are perpendicular if and onl if the have negative reciprocal slopes. That is to sa, if one line has slope m and the other has slope m then m m =. (You will be guided through a proof of this result in Eercise 7.) Please note that a horizontal line is perpendicular to a vertical line and vice versa, so we assume m 0 and m 0. In Eercises 65-70, ou are given a line and a point which is not on that line. Find the line perpendicular to the given line which passes through the given point. 65. = +, P (0, 0) 66. = 6 + 5, P (, ) 67. = 7, P (6, 0) 68. =, P (, ) 69. = 6, P (, ) 70. =, P ( 5, 0) 7. We shall now prove that = m + b is perpendicular to = m + b if and onl if m m =. To make our lives easier we shall assume that m > 0 and m < 0. We can also move the lines so that their point of intersection is the origin without messing things up, so we ll assume b = b = 0. (Take a moment with our classmates to discuss wh this is oka.) Graphing the lines and plotting the points O(0, 0), P (, m ) and Q(, m ) gives us the following set up. P O Q The line = m will be perpendicular to the line = m if and onl if OP Q is a right triangle. Let d be the distance from O to P, let d be the distance from O to Q and let d be the distance from P to Q. Use the Pthagorean Theorem to show that OP Q is a right triangle if and onl if m m = b showing d + d = d if and onl if m m =.

19 68 Linear and Quadratic Functions 7. Show that if a b, the line containing the points (a, b) and (b, a) is perpendicular to the line =. (Coupled with the result from Eample..7 on page, we have now shown that the line = is a perpendicular bisector of the line segment connecting (a, b) and (b, a). This means the points (a, b) and (b, a) are smmetric about the line =. We will revisit this smmetr in section 5..) 7. The function defined b I() = is called the Identit Function. (a) Discuss with our classmates wh this name makes sense. (b) Show that the point-slope form of a line (Equation.) can be obtained from I using a sequence of the transformations defined in Section.7.

20 . Linear Functions 69.. Answers. + = ( ) = 0. + = ( + 7) = 8. 8 = ( + 5) =. = ( + ) = = 5 ( 0) 6. = = ( + ) = = 0 = 7 9. = 5( ) = = ( 0) = 0. + = 678( + ) = = 5. =. = = = 5 6. = 8 7. = = = 0. =. f() = slope: m = -intercept: (0, ) -intercept: (, 0). f() = slope: m = -intercept: (0, ) -intercept: (, 0)

21 70 Linear and Quadratic Functions. f() = slope: m = 0 -intercept: (0, ) -intercept: none. f() = 0 slope: m = 0 -intercept: (0, 0) -intercept: {(, 0) is a real number} 5. f() = + slope: m = -intercept: ( 0, ) -intercept: (, 0) 6. f() = slope: m = -intercept: ( 0, ) -intercept: (, 0) 7. (, ) and ( 5, 7 ) 5 8. d(t) = t, t E(t) = 60t, t C() = 5 + 0, 0.. C(t) = 80t + 50, 0 t 8.. W () = , 0 She must make $5500 in weekl sales.. C(p) = 0.05p +.5 The slope 0.05 means it costs.5 per page. C(0) =.5 means there is a fied, or start-up, cost of $.50 to make each book.. F (m) =.5m +.05 The slope.5 means it costs an additional $.5 for each mile beond the first 0. miles. F (0) =.05, so according to the model, it would cost $.05 for a trip of 0 miles. Would this ever reall happen? Depends on the driver and the passenger, we suppose.

22 . Linear Functions 7 5. (a) F (C) = 9 5 C + (b) C(F ) = 5 9 (F ) = 5 9 F 60 9 (c) F ( 0) = 0 = C( 0). 6. N(T ) = 5 T + and N(0) = 5 howls per hour. Having a negative number of howls makes no sense and since N(07.5) = 0 we can put an upper bound of 07.5 F on the domain. The lower bound is trickier because there s nothing other than common sense to go on. As it gets colder, he howls more often. At some point it will either be so cold that he freezes to death or he s howling non-stop. So we re going to sa that he can withstand temperatures no lower than 60 F so that the applied domain is [ 60, 07.5]. { 6p +.5 if p 5 9. C(p) = 5.5p if p 6 { 5n if n 9 0. T (n) =.5n if n 0 { 0 if 0 m 500. C(m) = (m 500) if m > 500 { 0.c if c 00. P (c) = + 0.(c 00) if c > 00. (a) (b) (c) 8 if 0 d 5 D(d) = d + if 5 d 7 if 7 d 7 D(s) = if 0 s 0 s if 0 s 8 if s = D(d) 0 7 = D(s)

23 7 Linear and Quadratic Functions ( ) = ( ) = = h + h ( )( + h ) 5 5 = 5 ( ) ( ) = 0 (() + () 7) (( ) + ( ) 7) ( ) ( + h) h + 5. The average rate of change is h() h(0) 0 =. During the first two seconds after it is dropped, the object has fallen at an average rate of feet per second. (This is called the average velocit of the object.) F (8) F (0) 55. The average rate of change is 8 0 = 0.7. During the ears from 980 to 008, the average fuel econom of passenger cars in the US increased, on average, at a rate of 0.7 miles per gallon per ear. 56. (a) T () = 56, so at 0 AM ( hours after 6 AM), it is 56 F. T (8) = 6, so at PM (8 hours after 6 AM), it is 6 F. T () = 56, so at 6 PM ( hours after 6 AM), it is 56 F. T (8) T () (b) The average rate of change is 8 =. Between 0 AM and PM, the temperature increases, on average, at a rate of F per hour. T () T (8) (c) The average rate of change is 8 =. Between PM and 6 PM, the temperature decreases, on average, at a rate of F per hour. T () T () = 0. Between 0 AM and 6 PM, the tempera- (d) The average rate of change is ture, on average, remains constant. = 57. The average rate of change is C(5) C() 5 =. As production is increased from 000 to 5000 pens, the cost decreases at an average rate of $00 per 000 pens produced (0 per pen.) 59. = 60. = = 6. = 6. = 6. = = 66. = = = 69. = 70. = 0

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